Strange/Unexpected behavior of an Infinite product Some friends and I were playing around with this continued fraction:

We noticed when writing it out for each next step, the end behavior went either to 1 (when there was an even number of terms) or to a linear dependence on x (when there were odd). This was expected - but fitting the linear dependence gave something we didn't expect.
We wrote the odd-term fraction as a product:
.
We expected that the tail behavior would go as $(x+2m+2)$, but insead we found it closer to $(x+m+1)$: https://www.desmos.com/calculator/7ye4ijgoef
Its clear both give identical limits for finite $m$ as $x \longrightarrow \infty$, but
is there an argument as to why the significantly more accurate behavior $(x+m+1)$ occurs over the expected behavior $(x+2m+2)$ for large but fininte $x$?
 A: The limit you are looking for is $\lim_{x \to \infty}(f(x) - x)$. Write
$$\lim_{x \to \infty}(f(x) - x) = 2m+2 + \lim_{x \to \infty} x \cdot \left( \prod_{n=0}^m \frac{x+2n}{x+2n+1} - 1 \right) \,.$$
The limit in the right-hand side is not $0$, but rather $-m-1$. You can compute it as follows:
$$\begin{align*}\lim_{x \to \infty} x \cdot \left( \prod_{n=0}^m \frac{x+2n}{x+2n+1} - 1 \right)
&= \lim_{t \to 0} \frac1t \cdot \left( \prod_{n=0}^m \frac{\frac1t+2n}{\frac1t+2n+1} - 1 \right) \\
&= \left.\frac d{dt}\right\vert_{t=0} \prod_{n=0}^m \frac{1+2n t}{1+ (2n+1)t} \,. \end{align*}$$
This is easiest to compute using a Taylor expansion:
$$\begin{align*}\prod_{n=0}^m \frac{1+2n t}{1+ (2n+1)t}
&= \prod_{n=0}^m (1+2n t)(1- (2n+1)t + O(t^2)) \\
&= \prod_{n=0}^m (1-t + O(t^2)) \\
&= (1-t)^{m+1} + O(t^2) \end{align*}$$
so that our derivative is $-m-1$.
A: Transform the product to a sum using logarithms
$$P_m=\prod_{n=0}^m \frac{x+2 n}{x+2 n+1} \implies \log(P_m)=\sum_{n=0}^m \log \left(\frac{x+2 n}{x+2 n+1}\right)$$ Ubsing Taylor for large values of $x$
$$\log \left(\frac{x+2 n}{x+2 n+1}\right)=-\frac 1 x+\frac{4 n+1}{2 x^2}-\frac{12 n^2+6 n+1}{3 x^3}+O\left(\frac{1}{x^4}\right)$$ Summing from $n=0$ to $n=m$
$$\log(P_m)=-\frac{m+1}{x}+\frac{(m+1) (2 m+1)}{2 x^2}-\frac{(m+1)^2 (4 m+1)}{3 x^3}+O\left(\frac{1}{x^4}\right)$$
$$P_m=e^{\log(P_m)}=1-\frac{m+1}{x}+\frac{(m+1) (3 m+2)}{2 x^2}-\frac{(m+1)^2 (5 m+2)}{2 x^3}+O\left(\frac{1}{x^4}\right)$$
$$(x+2m+2)P_m=(x+m+1)-\frac{(m+1) (m+2)}{2 x}+\frac{(m+1)^2 (m+2)}{2
   x^2}+O\left(\frac{1}{x^3}\right)$$ which is even smaller that $(x+m+1)$.
Let us try with $x=10^3$ and $m=10$ the exact result is
$$\frac{4725919019374652598779904000}{4674801393661613629292257}=1010.934716$$
The truncated series gives
$$\frac{505467363}{500000}=1010.934726$$
Edit
Doing the same for a more general case, the first terms are
$$(x+am+b)P_m=\big[x +(a-1) m+(b-1)\big]-\frac{(m+1) ((2 a-3) m+2( b-1))}{2 x}$$
